Single-delete Nim

Nim is a well-known game that has been studied extensively over the years. In the most popular version of Nim, there are three piles of stones, and two players take turns. On their turn, a player selects one pile and removes any positive number of stones from that pile. The players continue taking turns, and the player who removes the last stone wins the game. This game can also be played with more than three piles following the same rules.

Nim is a two-player, zero-sum, finite, deterministic, and perfect information game with no possibility of a draw. Therefore, every configuration can be classified as either an N-position, where the player whose turn it is can force a win, or a P-position, where the player whose turn it is will inevitably lose if the opponent plays optimally. The classification of N-positions and P-positions in the aforementioned version of Nim was demonstrated by Bouton [3] in 1902. It became known that the conditions for the winning strategy in this game include mathematically intriguing structures.

Several variants of Nim have been proposed by modifying the rules of legal moves, such as Moore’s game [4] and Wythoff’s game [7]. The winning strategies of these variants have been studied extensively, and their mathematical formulations have attracted significant interest. In this study, we generalize Delete Nim, a game introduced by Abuku and Suetsugu [2], where players eliminate one pile and then split one of the remaining piles into two smaller piles in one move. We then investigate the winning strategy for this generalized version.

In this section, we first describe the rules of Delete Nim, which form the basis of this study. There are two variations of the Delete Nim rules provided in [2] and [6]. Although the rules differ in presentation, they are equivalent with respect to legal moves. In this paper, we adopt the latter rules¹¹1In [2], the rule described here is referred to as the “Variant of Delete Nim” (VDN)..

In Delete Nim, there are two piles of stones. On each turn, players perform the following two actions sequentially (we refer to these combined actions as a move), then pass the turn to their opponent.

• •

  Select one of the piles and remove all the stones from it, eliminating the pile.

• •

  Split the remaining pile into two new piles, ensuring that each new pile contains at least one stone.

A player unable to make this move on their turn loses the game. We denote a position in Delete Nim by $\langle y,z\rangle$, where $y$ and $z$ represent the number of stones in the two piles. Let $\mathbb{N}$ be the set of positive integers. Then, the set of all possible game positions (without specifying the current player’s turn) is denoted by $G_{2}=\{\,\langle y,z\rangle\,|\,y,z\in\mathbb{N}\}$. At the position $\langle 1,1\rangle$, the player whose turn it loses the game.

Similar to Nim, Delete Nim is a two-player, zero-sum, perfect information, deterministic, finite game with no possibility of a draw. Therefore, each position in $G_{2}$ can be classified as either an N-position or a P-position. The condition for determining this classification is as follows.

This theorem is included in Proposition 4.1 in this paper. It should be noted that in [2], not only is the above classification provided, but the Sprague-Grundy number of each position is also determined. In an N-position, a player can ensure winning by forcing their opponent into P-positions with optimal moves. For example, when the current position is $\langle 4,3\rangle$, the player on turn can proceed as $\langle 4,3\rangle\to\langle 3,1\rangle$ (and then the game proceeds as $\to\langle 2,1\rangle\to\langle 1,1\rangle$ and ends) and secure winning. Therefore, identifying the P-positions is equivalent to understanding a winning strategy and knowing the conditions for winning.

In this section, we define an extended rule that generalizes Delete Nim to any number of piles. Let $n$ be an integer greater than or equal to 2.

In Single-delete Nim, it is important to note that the number of piles remains $n$ at the end of each turn.

The following result for Single-delete Nim with three piles is known from Sakai [5]. To present the winning and losing conditions, we define the following notation: For $z\in\mathbb{N}$, if $z$ is divisible by $2^{k}$ but not divisible by $2^{k+1}$, we denote it as $v_{2}(z)=k$. When $z$ is odd, we define $v_{2}(z)=0$.

A proof of Theorem 1.4 is provided in [5] (pp.59-63). However, as it is also relevant for the case of four piles, we restate it in Section 2.1 of this paper.

As the main theorem of this paper, we present the necessary and sufficient condition to be a P-position in Single-delete Nim with four piles as Theorem 1.5. To state this theorem, we introduce notation related to the binary representation of positive integers. Let $I_{k}(z)$ represent the $k$-th digit from the right in the binary representation of $z\in\mathbb{N}$. In other words, if the quotient of $z$ divided by $2^{k-1}$ is even, then $I_{k}(z)=0$; if the quotient is odd, then $I_{k}(z)=1$. We note that $v_{2}(z)=\min\{\,k\,|\,I_{k}(z)=1\}-1$.

The organization of this paper is as follows. In Section 2, we present a proof of Theorem 1.4, which provides the classification of N-positions and P-positions for Single-delete Nim with three piles, along with several useful propositions. In Section 3, we give a proof of Theorem 1.5, which addresses the case with four piles. In Section 4, we offer a brief analysis of the scenario when the number of piles is five or more.

In this section, we present a detailed proof of Theorem 1.5, which establishes the necessary and sufficient conditions for a position in Single-delete Nim with four piles to qualify as a P-position. We first briefly explain the necessity of some conditions (1) to (5) in Theorem 1.5 through a comparison of specific examples. The necessity of all the conditions will be discussed in detail in the proof of Lemma 3.2 later.

1. (1).

   A specific example of a position in this case is $\langle w,x,y,z\rangle=\langle 1440,864,$
   $672,1120\rangle$. In its binary representation, it is

   $$\begin{array}[]{ll}w&10110100000\\ x&01101100000\\ y&01010100000\\ z&10001100000,\end{array}$$

   where the rightmost 1’s are in the same digits.

2. (2).

   A specific example of a position in this case is $\langle w,x,y,z\rangle=\langle 294,208,$
   $304,432\rangle$. In its binary representation, it is

   $$\begin{array}[]{ll}w&100100110\\ x&011010000\\ y&100110000\\ z&110110000.\end{array}$$

   In a similar position

   $$\begin{array}[]{ll}w&1001\underline{1}0110\\ x&011010000\\ y&100110000\\ z&110110000\end{array}$$

   where condition (2A) is not satisfied, by eliminating $x$ and partitioning $w$ into $w^{\prime}$ and $w^{\prime\prime}$, it is possible to fulfill condition (2) in a move, resulting in

   $$\begin{array}[]{ll}w^{\prime}&100100110\\ w^{\prime\prime}&000010000\\ y&100110000\\ z&110110000.\end{array}$$

3. (3).

   A specific example of a position in this case is $\langle w,x,y,z\rangle=\langle 669,468,$
   $800,288\rangle$. In its binary representation, it is

   $$\begin{array}[]{ll}w&1010011101\\ x&0111010100\\ y&1100100000\\ z&0100100000.\end{array}$$

   In a similar position

   $$\begin{array}[]{ll}w&10100\underline{0}1101\\ x&01110\underline{0}0100\\ y&1100100000\\ z&0100100000\end{array}$$

   where condition (3B) is not satisfied, by eliminating $z$ and partitioning $y$ into $y^{\prime}$ and $y^{\prime\prime}$, it is possible to fulfill condition (3) in a move, resulting in

   $$\begin{array}[]{ll}w&1010001101\\ x&0111000100\\ y^{\prime}&1100010000\\ y^{\prime\prime}&0000010000.\end{array}$$

4. (4).

   A specific example of a position in this case is $\langle w,x,y,z\rangle=\langle 11133,$
   $12716,7136,13312\rangle$. In its binary representation, it is

   $$\begin{array}[]{ll}w&10101101111101\\ x&11000110101100\\ y&01101111100000\\ z&11010000000000.\end{array}$$

   In a similar position

   $$\begin{array}[]{ll}w&10101101111101\\ x&11000110101100\\ y&011011\underline{0}1100000\\ z&11010000000000\end{array}$$

   where condition (4B) is not satisfied, by eliminating $x$ and partitioning $z$ into $z^{\prime}$ and $z^{\prime\prime}$, it is possible to fulfill condition (3) in a move, resulting in

   $$\begin{array}[]{ll}w&10101101111101\\ y&01101101100000\\ z^{\prime}&11001110000000\\ z^{\prime\prime}&00000010000000.\end{array}$$

5. (5).

   A specific example of a position in this case is $\langle w,x,y,z\rangle=\langle 45053,$
   $62932,32576,64512\rangle$. In its binary representation, it is

   $$\begin{array}[]{ll}w&1010111111111101\\ x&1111010111010100\\ y&0111111101000000\\ z&1111110000000000.\end{array}$$

   In a similar position

   $$\begin{array}[]{ll}w&1010111111111101\\ x&1111010111010100\\ y&011\underline{0}111101000000\\ z&1111110000000000\end{array}$$

   where condition (5A) is not satisfied, by eliminating $x$ and partitioning $z$ into $z^{\prime}$ and $z^{\prime\prime}$, it is possible to fulfill condition (4) in a move, resulting in

   $$\begin{array}[]{ll}w&1010111111111101\\ y&0110111101000000\\ z^{\prime}&1110110000000000\\ z^{\prime\prime}&0001000000000000.\end{array}$$

In this section, we will prove Theorem 1.5. Let $G_{4}=\{\,\langle w,x,y,z\rangle\,|\,w,x,y,z$
$\in\mathbb{N}\}$ be the set of all positions in Single-delete Nim with four piles. We say that $\langle w,x,y,z\rangle\in G_{4}$ is a standard form if it satisfies $v_{2}(w)\leq v_{2}(x)\leq v_{2}(y)\leq v_{2}(z)$. Rearranging the elements of $\langle w,x,y,z\rangle\in G_{4}$ in this order is referred to as the standardization of $\langle w,x,y,z\rangle$. We define the subset $P\subset G_{4}$ such that a position $\langle w,x,y,z\rangle$ belongs to $P$ if and only if its standardization satisfies any of the conditions (1) to (5) in Theorem 1.5. Let $N$ denote the set of all other positions in $G_{4}$. In other words, $P$ and $N$ correspond to the sets of P-positions and N-positions for this game, respectively.

In the progression of this game, the total sum of stones across all piles decreases monotonically, and since $P$ includes the terminal position $\langle 1,1,1,1\rangle$, the proof of Theorem 1.5 requires demonstrating the following two lemmas.

To prove these lemmas, we first define common notation. If the standard form of $\langle w,x,y,z\rangle$ satisfies condition ($i$) of Theorem 1.5 ($1\leq i\leq 5$), we denote $\langle w,x,y,z\rangle\in P_{i}$ and define the sets $P_{1},P_{2},P_{3},P_{4},P_{5}$ accordingly. Thus, $\displaystyle P=\bigcup_{i=1}^{5}P_{i}$, and by conditions (1) to (5) we have $P_{i}\cap P_{j}=\emptyset$ for any $i\neq j$.

We now present the following proposition to support our proof of the two lemmas.

Considering the carry in binary addition, it is clear that the above proposition holds. With this in mind, we will now proceed to prove Lemma 3.1 and Lemma 3.2.

Proof of Lemma 3.1. For a position $\langle w,x,y,z\rangle$ where $a=v_{2}(w),b=v_{2}(x),c=v_{2}(y),d=v_{2}(z)$, if any of the following conditions hold:

($\triangle$)  $a=b=c<d$, $a=b<c=d$, $a=b<c<d$, $a<b=c<d$

then none of the conditions (1) to (5) can be satisfied, and thus the position is not included in $P$. This fact will be frequently used in the explanation below. In this proof, when transitioning from $\langle w,x,y,z\rangle$ to $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle$ as a move, we show that if $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P$ then $\langle w,x,y,z\rangle\not\in P$. We consider the reverse of the operation:

• •

  First, select two of $w^{\prime},x^{\prime},y^{\prime},z^{\prime}$ and combine their sum into a single number.

• •

  Then, add a new element $u^{\prime}\in\mathbb{N}$.

We consider each case where $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{i}$ for $1\leq i\leq 5$.

1. a).

   In case $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{1}$:
   Since $v_{2}(w^{\prime})=v_{2}(x^{\prime})=v_{2}(y^{\prime})=v_{2}(z^{\prime})$, it is sufficient to consider only the case where the sum $y^{\prime}+z^{\prime}$ is formed in the reverse operation. In this situation, by Proposition 2.1(i), we have $v_{2}(w^{\prime})=v_{2}(x^{\prime})<v_{2}(y^{\prime}+z^{\prime})$. Therefore, regardless of the element $u^{\prime}$ added, $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle$ corresponds to ($\triangle$), and cannot be included in any $P_{i}$.

2. b).

   In case $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{2}$:
   Suppose $v_{2}(w^{\prime})<v_{2}(x^{\prime})=v_{2}(y^{\prime})=v_{2}(z^{\prime})$. Let $b=v_{2}(x^{\prime})$, and we will use the following facts:
   ($\clubsuit_{1}$) From (2A), we have $I_{b+1}(w^{\prime})=0$.
   ($\clubsuit_{2}$) $I_{b+1}(w^{\prime}+z^{\prime})=1$?D

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle$, formed by the reverse operation where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime}+z^{\prime})$.

     • –

       If $v_{2}(u^{\prime})>v_{2}(x^{\prime})$, then by ($\clubsuit_{1}$), it cannot satisfy (3C), (4E) or (5F).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(w^{\prime})$, it corresponds to ($\triangle$).

     • –

       If $v_{2}(u^{\prime})<v_{2}(x^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\clubsuit_{1}$), it cannot satisfy (4C) or (5D).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle\not\in P$.

   • •

     We consider whether $\langle u^{\prime},w^{\prime}+z^{\prime},x^{\prime},y^{\prime}\rangle$, formed by the reverse operation where the sum $w^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime}+z^{\prime})<v_{2}(x^{\prime})=v_{2}(y^{\prime})$.

     • –

       If $v_{2}(u^{\prime})>v_{2}(y^{\prime})$, it corresponds to ($\triangle$).

     • –

       If $v_{2}(u^{\prime})\leq v_{2}(y^{\prime})$, then by ($\clubsuit_{2}$), it cannot satisfy (2A) or (3A).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime}+z^{\prime},x^{\prime},y^{\prime}\rangle\not\in P$.

3. c).

   In case $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{3}$:
   Suppose $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime})=v_{2}(z^{\prime})$. Let $c=v_{2}(y^{\prime})$, and we will use the following facts:
   ($\diamondsuit_{1}$) From (3A), we have $I_{c+1}(w^{\prime})=I_{c+1}(x^{\prime})=0$.
   ($\diamondsuit_{2}$) $I_{c+1}(x^{\prime}+z^{\prime})=1$.
   ($\diamondsuit_{3}$) From (3B) and Proposition 3.3, we have $I_{c+1}(w^{\prime}+x^{\prime})=1$.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle$, formed by the reverse operation where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime})<c<v_{2}(y^{\prime}+z^{\prime})$.

     • –

       If $v_{2}(u^{\prime})>c$, then by ($\diamondsuit_{1}$), it cannot satisfy (3B), (4D) or (5E).

     • –

       If $v_{2}(u^{\prime})=c$, then by ($\diamondsuit_{1}$), it cannot satisfy (4C) or (5D).

     • –

       If $v_{2}(u^{\prime})<c$, $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(x^{\prime})$, then by ($\diamondsuit_{1}$), it cannot satisfy (4B) or (5C).

     • –

       If $v_{2}(u^{\prime})=v_{2}(w^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(x^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle\not\in P$.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime}+z^{\prime},y^{\prime}\rangle$, formed by the reverse operation where the sum $x^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime}+z^{\prime})<v_{2}(y^{\prime})$.

     • –

       If $v_{2}(u^{\prime})>v_{2}(y^{\prime})$, then by ($\diamondsuit_{1}$), it cannot satisfy (4C) or (5D).

     • –

       If $v_{2}(u^{\prime})=v_{2}(y^{\prime})$, then by ($\diamondsuit_{2}$), it cannot satisfy (3A).

     • –

       If $v_{2}(u^{\prime})<v_{2}(y^{\prime})$, $v_{2}(u^{\prime})\neq v_{2}(x^{\prime}+z^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\diamondsuit_{1}$) and ($\diamondsuit_{2}$), it cannot satisfy (4A) or (5B).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime}+w^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(z^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime}+z^{\prime},y^{\prime}\rangle\not\in P$.

   • •

     For $\langle u^{\prime},w^{\prime}+z^{\prime},x^{\prime},y^{\prime}\rangle$, formed by the reverse operation where the sum $w^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, it can similarly be shown that it is not included in $P$, just as in the case where the sum $x^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added.

   • •

     We consider whether $\langle u^{\prime},w^{\prime}+x^{\prime},y^{\prime},z^{\prime}\rangle$, formed by the reverse operation where the sum $w^{\prime}+x^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime}+x^{\prime})<v_{2}(y^{\prime})=v_{2}(z^{\prime})$.

     • –

       If $v_{2}(u^{\prime})>v_{2}(y^{\prime})$, it corresponds to ($\triangle$).

     • –

       If $v_{2}(u^{\prime})\leq v_{2}(y^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime}+x^{\prime})$, then by ($\diamondsuit_{3}$), it cannot satisfy (2A) or (3A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(w^{\prime}+x^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime}+x^{\prime},y^{\prime},z^{\prime}\rangle\not\in P$.

4. d).

   In case $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{4}$:
   Suppose $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime})<v_{2}(z^{\prime})$. Let $c=v_{2}(y^{\prime}),d=v_{2}(z^{\prime})$, and we will use the following facts:
   ($\heartsuit_{1}$) From (4A), we have $I_{d+1}(w^{\prime})=I_{d+1}(x^{\prime})=I_{d+1}(y^{\prime})=0$.
   ($\heartsuit_{2}$) From (4C), we have $I_{c+1}(w^{\prime})=I_{c+1}(x^{\prime})=1$.
   ($\heartsuit_{3}$) From (4B), we have $I_{j}(w)+I_{j}(x)+I_{j}(y)\geq 2$ for $c+2\leq j\leq d$.
   ($\heartsuit_{4}$) From (4B) and by using similar arguments as Proposition 3.3,
      we have $I_{d+1}(w^{\prime}+x^{\prime})=I_{d+1}(w^{\prime}+y^{\prime})=I_{d+1}(x^{\prime}+y^{\prime})=1$.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle$, formed by the reverse operation where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime}+z^{\prime})<d$.

     • –

       If $v_{2}(u^{\prime})>d$, then by ($\heartsuit_{1}$), it cannot satisfy (4B) or (5C).

     • –

       If $v_{2}(u^{\prime})=d$, then by ($\heartsuit_{1}$), it cannot satisfy (4A) or (5B).

     • –

       If $v_{2}(u^{\prime})<d$, $v_{2}(u^{\prime})\neq v_{2}(y^{\prime}+z^{\prime})$, $v_{2}(u^{\prime})\neq v_{2}(x^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\heartsuit_{1}$) and ($\heartsuit_{3}$), it cannot satisfy (4A) or (5A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(y^{\prime}+z^{\prime})$, then by ($\heartsuit_{2}$), it cannot satisfy (3A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(w^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle\not\in P$.

   • •

     For $\langle u^{\prime},w^{\prime},x^{\prime}+z^{\prime},y^{\prime}\rangle$, formed by the reverse operation with the sum $x^{\prime}+z^{\prime}$ and addition of $u^{\prime}$, and for $\langle u^{\prime},w^{\prime}+z^{\prime},x^{\prime},y^{\prime}\rangle$, similarly with the sum $w^{\prime}+z^{\prime}$, it can be shown that neither is included in $P$, just as in the case where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime}+y^{\prime},z^{\prime}\rangle$, formed by the reverse operation where the sum $x^{\prime}+y^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime}+y^{\prime})<v_{2}(z^{\prime})=d$.

     • –

       If $v_{2}(u^{\prime})>v_{2}(z^{\prime})$, then by ($\heartsuit_{1}$), it cannot satisfy (4C) or (5D).

     • –

       If $v_{2}(u^{\prime})=v_{2}(z^{\prime})$, then by ($\heartsuit_{4}$), it cannot satisfy (3A).

     • –

       If $v_{2}(u^{\prime})<v_{2}(z^{\prime})$, $v_{2}(u^{\prime})\neq v_{2}(x^{\prime}+y^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\heartsuit_{1}$) and ($\heartsuit_{4}$), it cannot satisfy (4A) or (5B).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime}+y^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(w^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime}+y^{\prime},z^{\prime}\rangle\not\in P$.

   • •

     For $\langle u^{\prime},w^{\prime}+y^{\prime},x^{\prime},z^{\prime}\rangle$, formed by the reverse operation with the sum $w^{\prime}+y^{\prime}$ and addition of $u^{\prime}$, and for $\langle u^{\prime},w^{\prime}+x^{\prime},y^{\prime},z^{\prime}\rangle$, similarly with the sum $w^{\prime}+x^{\prime}$, it can be shown that neither is included in $P$, just as in the case where the sum $x^{\prime}+y^{\prime}$ is taken and $u^{\prime}$ is added.

5. e).

   In case $\langle w^{\prime},x^{\prime},y^{\prime},z^{\prime}\rangle\in P_{5}$:
   Suppose $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime})<v_{2}(z^{\prime})$. Let $b=v_{2}(x^{\prime}),c=v_{2}(y^{\prime}),d=v_{2}(z^{\prime})$, and define the smallest $e$ such that $e>d$ and
   ($\spadesuit_{1}$) $I_{e+1}(w^{\prime})=I_{e+1}(x^{\prime})=I_{e+1}(y^{\prime})=I_{e+1}(z^{\prime})=0$.
   We will use the following facts:
   ($\spadesuit_{2}$) From (5A), we have $I_{i}(w^{\prime})+I_{i}(x^{\prime})+I_{i}(y^{\prime})+I_{i}(z^{\prime})\geq 3$ for
     $d+2\leq i\leq e$.
   ($\spadesuit_{3}$) From (5B), we have $I_{d+1}(w^{\prime})=I_{d+1}(x^{\prime})=I_{d+1}(y^{\prime})=1$ and
      $I_{d+1}(w^{\prime}+z^{\prime})=I_{d+1}(x^{\prime}+z^{\prime})=I_{d+1}(y^{\prime}+z^{\prime})=0$.
   ($\spadesuit_{4}$) From (5D), we have $I_{c+1}(w^{\prime})=I_{c+1}(x^{\prime})=1$.
   ($\spadesuit_{5}$) From (5C), we have $I_{j}(w)+I_{j}(x)+I_{j}(y)\geq 2$ for $c+2\leq j\leq d$,
     and from (5E), $I_{k}(w)+I_{k}(x)\geq 1$ for $b+2\leq k\leq c$.
   ($\spadesuit_{6}$) From (5A) and by using similar arguments as Proposition 3.3, regarding the sum of any two of $w^{\prime},x^{\prime},y^{\prime},z^{\prime}$, $I_{e+1}(w^{\prime}+x^{\prime})=\cdots=I_{e+1}(y^{\prime}+z^{\prime})=1$.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle$, formed by the reverse operation where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime})<v_{2}(y^{\prime}+z^{\prime})=c$.

     • –

       If $v_{2}(u^{\prime})>e$, then by ($\spadesuit_{1}$), it cannot satisfy (4B).

     • –

       If $d<v_{2}(u^{\prime})\leq e$, then by ($\spadesuit_{2}$) it cannot satisfy (4A) , and by ($\spadesuit_{1}$) and ($\spadesuit_{6}$) it cannot satisfy (5A).

     • –

       If $v_{2}(u^{\prime})=d$, then by ($\spadesuit_{3}$), it cannot satisfy (4A) and (5B).

     • –

       If $v_{2}(u^{\prime})<d$, $v_{2}(u^{\prime})\neq v_{2}(y^{\prime}+z^{\prime})$, $v_{2}(u^{\prime})\neq v_{2}(x^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\spadesuit_{5}$) it cannot satisfy (4A), and by ($\spadesuit_{1}$) and ($\spadesuit_{6}$) it cannot satisfy (5A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(y^{\prime}+z^{\prime})$, then by ($\spadesuit_{4}$), it cannot satisfy (3A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(w^{\prime})$, it corresponds to ($\triangle$).

     Therefore we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime},y^{\prime}+z^{\prime}\rangle\not\in P$.

   • •

     For $\langle u^{\prime},w^{\prime},x^{\prime}+z^{\prime},y^{\prime}\rangle$, formed by the reverse operation with the sum $x^{\prime}+z^{\prime}$ and addition of $u^{\prime}$, and for $\langle u^{\prime},w^{\prime}+z^{\prime},x^{\prime},y^{\prime}\rangle$, similarly with the sum $w^{\prime}+z^{\prime}$, it can be shown that neither is included in $P$, just as in the case where the sum $y^{\prime}+z^{\prime}$ is taken and $u^{\prime}$ is added.

   • •

     We consider whether $\langle u^{\prime},w^{\prime},x^{\prime}+y^{\prime},z^{\prime}\rangle$, formed by the reverse operation where the sum $x^{\prime}+y^{\prime}$ is taken and $u^{\prime}$ is added, is included in $P$. Here, we have $v_{2}(w^{\prime})<v_{2}(x^{\prime}+y^{\prime})<v_{2}(z^{\prime})=d$.

     • –

       If $v_{2}(u^{\prime})>e$, then by ($\spadesuit_{1}$), it cannot satisfy (4B).

     • –

       If $v_{2}(z^{\prime})<v_{2}(u^{\prime})\leq e$, then by ($\spadesuit_{2}$) it cannot satisfy (4A), and by ($\spadesuit_{1}$) and ($\spadesuit_{6}$) it cannot satisfy (5A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(z^{\prime})$, then by ($\spadesuit_{3}$), it cannot satisfy (3A).

     • –

       If $v_{2}(u^{\prime})<v_{2}(z^{\prime})$, $v_{2}(u^{\prime})\neq v_{2}(x^{\prime}+y^{\prime})$ and $v_{2}(u^{\prime})\neq v_{2}(w^{\prime})$, then by ($\spadesuit_{5}$) it cannot satisfy (4A) and by ($\spadesuit_{1}$) and ($\spadesuit_{6}$) it cannot satisfy (5A).

     • –

       If $v_{2}(u^{\prime})=v_{2}(x^{\prime}+y^{\prime})$ or $v_{2}(u^{\prime})=v_{2}(w^{\prime})$, it corresponds to ($\triangle$).

     Therefore, we can conclude that $\langle u^{\prime},w^{\prime},x^{\prime}+y^{\prime},z^{\prime}\rangle\not\in P$.

   • •

     For $\langle u^{\prime},w^{\prime}+y^{\prime},x^{\prime},z^{\prime}\rangle$, formed by the reverse operation with the sum $w^{\prime}+y^{\prime}$ and addition of $u^{\prime}$, and for $\langle u^{\prime},w^{\prime}+x^{\prime},y^{\prime},z^{\prime}\rangle$, similarly with the sum $w^{\prime}+x^{\prime}$, it can be shown that neither is included in $P$, just as in the case where the sum $x^{\prime}+y^{\prime}$ is taken and $u^{\prime}$ is added.

Thus we completes the proof. $\Box$